Problem: Let $g(x)=-4\cdot3^x$. Find $g'(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $-4\log_3(x)\cdot 3^x$ (Choice B) B $-4\ln(x)\cdot 3^x$ (Choice C) C $-4\ln(3)\cdot 3^x$ (Choice D) D $-4\cdot 3^{x-1}$
Solution: The expression for $g(x)$ includes an exponential term. Remember that the derivative of the general exponential term $a^x$ (where $a$ is any positive constant) is $\ln(a)\cdot a^x$. Put another way, $\dfrac{d}{dx}(a^x)=\ln(a)\cdot a^x$. $\begin{aligned} g'(x)&=\dfrac{d}{dx}(-4\cdot3^x) \\\\ &=-4\dfrac{d}{dx}(3^x) \\\\ &=-4\cdot\ln(3)\cdot3^x \\\\ \end{aligned}$ In conclusion, $g'(x)=-4\ln(3)\cdot3^x$.